A189937 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sin(Pi/8), t=cos(Pi/8).

1, 3, 6, 8, 10, 13, 15, 18, 20, 22, 25, 27, 29, 31, 33, 36, 38, 40, 43, 45, 48, 50, 52, 55, 57, 59, 61, 63, 66, 68, 70, 73, 75, 78, 80, 82, 85, 87, 89, 91, 93, 96, 98, 100, 103, 105, 107, 110, 112, 115, 117, 119, 121, 123, 126, 128, 130, 133, 135, 137, 140, 142, 145, 147, 149, 151, 153, 156, 158, 160, 163, 165, 167, 170, 172, 175, 177

Offset

1,2

Comments

This is one of three sequences that partition the positive integers. In general, suppose that r, s, t are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define b(n) and c(n) as the ranks of n/s and n/t. It is easy to prove that

a(n) = n + [n*s/r] + [n*t/r],

b(n) = n + [n*r/s] + [n*t/s],

c(n) = n + [n*r/t] + [n*s/t], where []=floor.

Taking r=1, s=sin(Pi/8), t=cos(Pi/8) gives a=A189937, b=A189938, c=A189939.

Links

G. C. Greubel, Table of n, a(n) for n = 1..10000

Formula

A189937: a(n) = n + [n*sin(Pi/8)] + [n*cos(Pi/8)].

A189938: b(n) = n + [n*csc(Pi/8)] + [n*cot(Pi/8)].

A189939: c(n) = n + [n*sec(Pi/8)] + [n*tan(Pi/8)].

Mathematica

r=1; s=Sin[Pi/8]; t=Cos[Pi/8];
a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
c[n_] := n + Floor[n*r/t] + Floor[n*s/t];
Table[a[n], {n, 1, 120}]  (*A189937*)
Table[b[n], {n, 1, 120}]  (*A189938*)
Table[c[n], {n, 1, 120}]  (*A189939*)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*sin(Pi/8)) + floor(n*cos(Pi/8)), ", ")) \\ G. C. Greubel, Jan 13 2018
(Magma) C<i> := ComplexField(); [n + Floor(n*Sin(Pi(C)/8)) + Floor(n*Cos(Pi(C)/8)): n in [1..100]]; // G. C. Greubel, Jan 13 2018

Crossrefs

Cf. A189938, A189939.

Sequence in context: A304500 A047282 A304497 * A190325 A064437 A287180

Adjacent sequences: A189934 A189935 A189936 * A189938 A189939 A189940

Keyword

nonn

Author

Clark Kimberling, May 01 2011

Status

approved